4
$\begingroup$

Consider a real Klein-Gordon field $\phi$ in a globally hyperbolic spacetime, with metric $g_{\mu\nu}$.

The covariant Klein-Gordon equation is

$$(g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}+m^{2})\phi=0$$

Let $V$ be the vector space of all real-valued solutions to this equation.

Define the following bilinear form:

$$(f,h):=\int_{\Sigma} d\Sigma_{\mu}\sqrt g g^{\mu\nu} (f\partial_{\nu}h-h\partial_{\nu}f), \forall f,h \in V$$

where $\Sigma$ is any spacelike hypersurface.

This is a bilinear, antisymmetric, non-degenerate form, i.e. a symplectic form.

How can I prove that it is non-degenerate?

$\endgroup$
3
  • $\begingroup$ Probably some kind of bump function. If f is nonzero somewhere, try picking some h which is 1 nearby and 0 elsewhere. Probably have to be a bit more careful since there are derivatives involved, try integrating the first bit by parts maybe? $\endgroup$ Commented Jan 4 at 16:16
  • $\begingroup$ I thought maybe you can find a set of basis $\{f_1,...,f_n\}$, and rewrite $f$ and $h$ by $f=a_1 f_1+...+a_n f_n$, $h=b_1 f_1+...+b_n f_n$ (it could be non-discrete), and then show that only when $a_1=a_2=...=a_n=0$ or $b_1=b_2=...=b_n=0$, the inner product could be zero. $\endgroup$
    – David Shaw
    Commented Jan 5 at 4:12
  • $\begingroup$ In this particular case, if $f=0$ whenever $(f,h) = 0$ for all of $h$, then the inner product you wrote down is nondegenerate. This is the usual definition, so I am more than likely missing some details since it is bilinear (like something about isomorphisms between vector spaces) but this is the bare bone version. $\endgroup$
    – MathZilla
    Commented Jan 5 at 16:15

1 Answer 1

1
$\begingroup$

It is easy actually. To be precise $\Sigma$ is a smooth spacelike Cauchy surface of the spacetime and the considered space $V$ of solutions of the KG equations is made of solutions smooth and with compactly supported Cauchy data on $\Sigma$. Under these hypotheses, the relation between Cauchy data on $\Sigma$, $(f|_\Sigma, n_\Sigma \cdot\nabla f|_\Sigma)$ and corresponding solutions $f$ of the KG equation is one-to-one. In other words the Cauchy problem is well posed.

If $(f,h)=0$ for every $h$ then both $f$ and its derivative normal to $\Sigma$ are zero. This easily follows from the very form of the simplectic form and from the fact that we can choose the Cauchy data of $h$ arbitrarily and there is a corresponding $h$.

Hence, again in view of the well posedness of the Cauchy problem, $f$ must be the zero solution of the KG equation.

$\endgroup$
2
  • $\begingroup$ $(f,h)=0, \forall h$ implies, for the well posedness, that the integral is $0$ for every possible pair $(h|_\Sigma, n_\Sigma \cdot\nabla h|_\Sigma)$, which implies that $(f|_\Sigma, n_\Sigma \cdot\nabla f|_\Sigma)=(0,0)$, which implies, for the well posedness, that $f=0$. Right? $\endgroup$
    – Ric
    Commented Jan 7 at 14:06
  • $\begingroup$ Yes, right. You understood $\endgroup$ Commented Jan 7 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.